A speeder is driving down the road at a constant 20 m/s, he passes a patrolman parked on the roadside. The patrolman waits 3 seconds, then pursues the speeder, accelerating at a constant 4.0 m/s. How long does it take the patrolman to catch the speeder? How far has he traveled before doing so?
Can you PLEASE explain how you got this answer?
To find out how long it takes for the patrolman to catch the speeder, we can set up the following equations:
For the speeder:
Distance traveled by the speeder (Ds) = Speed of the speeder (Vs) × Time taken (Ts)
For the patrolman:
Distance traveled by the patrolman (Dp) = Initial velocity of the patrolman (Vp) × Time taken (Tp) + 0.5 × Acceleration of the patrolman (Ap) × (Time taken (Tp))^2
Now, let's solve for the time taken (Ts) for the speeder to be caught by the patrolman.
Let the time taken by the patrolman to catch the speeder be Tc.
Since the patrolman waits for 3 seconds before pursuing, the time taken by the patrolman is given by:
Tp = Ts + 3
Since the speed of the patrolman is constant, we can rewrite the equation for the distance traveled by the patrolman as:
Dp = Speed of the patrolman × Time taken by the patrolman
Dp = Vp × Tp
Now, let's substitute the value of Tp:
Dp = Vp × (Ts + 3)
Since the patrolman catches the speeder when the distances traveled by both are equal, we can equate the distance traveled by the speeder to the distance traveled by the patrolman:
Ds = Dp
Substituting the equations for the distances traveled:
Vs × Ts = Vp × (Ts + 3)
Now, let's solve for Ts:
Vs × Ts = Vp × Ts + Vp × 3
Vs × Ts - Vp × Ts = Vp × 3
Ts × (Vs - Vp) = Vp × 3
Ts = Vp × 3 / (Vs - Vp)
Given:
Vs = 20 m/s (Speed of the speeder)
Vp = 0 m/s (Initial velocity of the patrolman before acceleration)
Ap = 4.0 m/s^2 (Acceleration of the patrolman)
Substituting the values:
Ts = (0 × 3) / (20 - 0)
Ts = 0 / 20
Ts = 0
This means the speeder is caught by the patrolman instantaneously without any time delay.
To find the distance traveled before being caught, we can substitute the value of Ts in the distance equation for the speeder:
Ds = Vs × Ts
Ds = 20 × 0
Ds = 0 meters
Therefore, it takes the patrolman 0 seconds to catch the speeder, and the patrolman has traveled 0 meters before doing so.
To solve this problem, we can use the equations of motion to find the time it takes for the patrolman to catch the speeder and the distance traveled by the patrolman before catching the speeder.
Let's define the following variables:
- t = time taken by the patrolman to catch the speeder
- d = distance traveled by the patrolman before catching the speeder
We know that the speeder is driving at a constant speed of 20 m/s, and the patrolman accelerates at a constant rate of 4.0 m/s².
For the speeder:
- Speed (v1) = 20 m/s
For the patrolman:
- Initial velocity (u2) = 0 m/s (since the patrolman is parked)
- Acceleration (a2) = 4.0 m/s²
- Time (t) = ?
To find the time taken by the patrolman to catch the speeder, we can use the equation of motion:
v2 = u2 + a2t
Since the patrolman catches up with the speeder, their final velocities will be equal. Thus:
v2 = v1
Substituting the values into the equation, we get:
v1 = u2 + a2t
20 m/s = 0 m/s + (4.0 m/s²)t
Simplifying the equation, we find:
(4.0 m/s²)t = 20 m/s
t = 20 m/s / 4.0 m/s²
t = 5 seconds
Therefore, it takes the patrolman 5 seconds to catch the speeder.
To find the distance traveled by the patrolman before catching the speeder, we can use the equation of motion:
d = u2t + 0.5a2t²
Substituting the values into the equation, we get:
d = (0 m/s)(5 s) + 0.5(4.0 m/s²)(5 s)²
d = 0 m + 0.5(4.0 m/s²)(25 s²)
d = 0 + 0.5(100 m)
d = 50 meters
Therefore, the patrolman travels a distance of 50 meters before catching the speeder.
s=v(t+3)
s=at²/2
v(t+3) =at²/2
at²-2vt-6v=0
t²-10t-30=0
t=12.42 s
s= v(t+3)=20(12.42+3) = 308.4 m